If overline{a}=2 hat{i}+3 hat{j}+2 hat{k},ove | vedclass

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(C) Let $\overline{d} = \overline{a} + \lambda \overline{b}$.$\overline{d} = (2 \hat{i} + 3 \hat{j} + 2 \hat{k}) + \lambda(2 \hat{i} + \hat{j} - \hat{

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$\frac{-11}{5}$

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(C) Let $\overline{d} = \overline{a} + \lambda \overline{b}$.$\overline{d} = (2 \hat{i} + 3 \hat{j} + 2 \hat{k}) + \lambda(2 \hat{i} + \hat{j} - \hat{k})$$\overline{d} = (2 + 2\lambda) \hat{i} + (3 + \lambda) \hat{j} + (2 - \lambda) \hat{k}$.Since $\overline{d}$ is perpendicular to $\overline{c}$,their dot product must be zero,i.e.,$\overline{c} \cdot \overline{d} = 0$.$(\hat{i} + 3 \hat{j}) \cdot [(2 + 2\lambda) \hat{i} + (3 + \lambda) \hat{j} + (2 - \lambda) \hat{k}] = 0$.$1(2 + 2\lambda) + 3(3 + \lambda) + 0(2 - \lambda) = 0$.$2 + 2\lambda + 9 + 3\lambda = 0$.$5\lambda + 11 = 0$.$\lambda = -\frac{11}{5}$.

If $\overline{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overline{c}=\hat{i}+3 \hat{j}$ are such that $(\overline{a}+\lambda \overline{b})$ is perpendicular to $\overline{c}$,then the value of $\lambda$ is

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